A mixture of `CaCl_(2)` and NaCl weighing 4.44 is treated with sodium carbonate solution to precipitate all the `Ca^(2+)` ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to get 0.56 g of `CaO`. The percentage of NaCl in the mixture of (atomic mass of Ca=40) is

Options

A. `75`
B. `30.6`
C. `25`
D. `69.4`

Explanation

To solve the problem, we need to determine the percentage of NaCl in a mixture of CaCl₂ and NaCl weighing 4.44 grams. The process involves several steps, including calculating the amount of CaCl₂ based on the mass of CaO produced after heating the calcium carbonate.

### Step-by-Step Solution:

1. **Determine the mass of CaO produced**:
We know that the mass of CaO obtained after heating calcium carbonate is 0.56 g.

2. **Calculate the moles of CaO**:
The molar mass of CaO (Calcium = 40 g/mol, Oxygen = 16 g/mol) is:
\[ \text{Molar mass of CaO} = 40 + 16 = 56 \text{ g/mol} \]
Now, we can calculate the moles of CaO produced:
\[ \text{Moles of CaO} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.56 \text{ g}}{56 \text{ g/mol}} = 0.01 \text{ mol} \]

3. **Determine the moles of CaCO₃ produced**:
The reaction for the formation of CaO from CaCO₃ is:
\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]
From the stoichiometry of the reaction, 1 mole of CaCO₃ produces 1 mole of CaO. Therefore, the moles of CaCO₃ produced is also 0.01 mol.

4. **Calculate the mass of CaCO₃**:
The molar mass of CaCO₃ (Calcium = 40 g/mol, Carbon = 12 g/mol, Oxygen = 16 g/mol × 3) is:
\[ \text{Molar mass of CaCO}_3 = 40 + 12 + (16 \times 3) = 100 \text{ g/mol} \]
Now, we can calculate the mass of CaCO₃ produced:
\[ \text{Mass of CaCO}_3 = \text{moles} \times \text{molar mass} = 0.01 \text{ mol} \times 100 \text{ g/mol} = 1.0 \text{ g} \]

5. **Determine the moles of CaCl₂ in the mixture**:
The reaction between CaCl₂ and Na₂CO₃ produces CaCO₃ and NaCl:
\[ \text{CaCl}_2 + \text{Na}_2\text{CO}_3 \rightarrow \text{CaCO}_3 + 2\text{NaCl} \]
The molar mass of CaCl₂ is:
\[ \text{Molar mass of CaCl}_2 = 40 + (35.5 \times 2) = 111 \text{ g/mol} \]
Since 1 mole of CaCl₂ produces 1 mole of CaCO₃, the moles of CaCl₂ used to produce 1.0 g of CaCO₃ is:
\[ \text{Moles of CaCl}_2 = 0.01 \text{ mol} \]

6. **Calculate the mass of CaCl₂**:
\[ \text{Mass of CaCl}_2 = \text{moles} \times \text{molar mass} = 0.01 \text{ mol} \times 111 \text{ g/mol} = 1.11 \text{ g} \]

7. **Determine the mass of NaCl in the mixture**:
The total mass of the mixture is given as 4.44 g. Therefore, the mass of NaCl can be calculated as:
\[ \text{Mass of NaCl} = \text{Total mass} - \text{Mass of CaCl}_2 = 4.44 \text{ g} - 1.11 \text{ g} = 3.33 \text{ g} \]

8. **Calculate the percentage of NaCl in the mixture**:
\[ \text{Percentage of NaCl} = \left(\frac{\text{Mass of NaCl}}{\text{Total mass}}\right) \times 100 = \left(\frac{3.33 \text{ g}}{4.44 \text{ g}}\right) \times 100 \approx 75\% \]

### Final Answer:
The percentage of NaCl in the mixture is approximately **75%**.